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Question Number 156270 by Tawa11 last updated on 09/Oct/21

$$\mathrm{Show}\mathrm{that}\mathrm{i}={\sin }^{-1}{\left(\frac{\sqrt{2}}{2}\right)}^2\mathrm{in}\mathrm{air},\mathrm{if}\mathrm{the}\mathrm{refractive}$$$$\mathrm{index}\mathrm{n}=\frac{{\sin }^2\left(60\right)}{{\sin }^2\left(45\right)}$$

Commented by Tawa11 last updated on 10/Oct/21

$$\mathrm{Sir},\mathrm{I}\mathrm{just}\mathrm{know}\mathrm{that},\mathrm{refractive}\mathrm{index},\mathrm{n}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}$$

Answered by ajfour last updated on 10/Oct/21

$${\mathrm{n}}_{\mathrm{air}}\sin \mathrm{i}={\mathrm{n}}_{\mathrm{medium}}\sin \mathrm{r}$$$$\sin \mathrm{i}=\frac{{\mathrm{n}}_{\mathrm{medium}}}{{\mathrm{n}}_{\mathrm{air}}}\sin \mathrm{r}$$$$=\mathrm{nsin}\mathrm{r}=\frac{{\sin }^2\left(60\right)}{{\sin }^2\left(45\right)}\sin \mathrm{r}$$$$\mathrm{i}={\sin }^{-1}\left\{\frac{{\sin }^2\left(60\right)}{{\sin }^2\left(45\right)}\sin \mathrm{r}\right\}$$$$$$

Commented by Tawa11 last updated on 11/Oct/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

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